Ramanujan's 6-10-8 Identity turns out to depend on a special case of, $$u_1^k+u_2^k+u_3^k =v_1^k+v_2^k+v_3^k$$ simultaneously valid for $k=2,4$. I was investigating if the next system $k=2,6$, $$x_1^2+x_2^2+x_3^2 =y_1^2+y_2^2+y_3^2\\x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6\tag1$$ would have something similar. I observed empirically that, $$\left(\sum_{i=1}^3\big(x_i^{10}-y_i^{10}\big)\right)\left(\sum_{i=1}^3\big(x_i^{4}-y_i^{4}\big)\right)^2=20\prod_{i=1}^3\prod_{j=1}^3\big(x_i^2-y_j^2\big)\tag2$$ Example: $$10^k+15^k+23^k = 3^k+19^k+22^k$$ yields, $$\small \text{LHS}= \big(10^{10} + 15^{10} + 23^{10} - 3^{10} - 19^{10} - 22^{10}\big)\big(10^4 + 15^4 + 23^4 - 3^4 - 19^4 - 22^4\big)^2$$ $$\small \text{RHS}=20(10^2 - 3^2)(10^2 - 19^2)(10^2 - 22^2)(15^2 - 3^2)(15^2 - 19^2)(15^2 - 22^2)(23^2 - 3^2)(23^2 - 19^2)(23^2 - 22^2)$$ $$\small\text{LHS}=\text{RHS}=37739520^2\times3830610$$ I've also tested it with more general parametric solutions and it works just fine.

Q: But how do we prove $(2)$ rigorously?

  • $\begingroup$ why so many squares ? $\endgroup$
    – mercio
    Commented Dec 11, 2016 at 14:34
  • $\begingroup$ @mercio: I observed this while working with $x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k$ for $k=2,4$. I'm not sure if the version $k=1,3$ and signed $x_i, y_i$ will work, but fortunately $(1)$ has infinitely many solutions to play with. $\endgroup$ Commented Dec 11, 2016 at 14:38
  • $\begingroup$ @Tito Piezas III Is this question still of interest? $\endgroup$
    – Old Peter
    Commented Jan 27, 2017 at 15:00
  • $\begingroup$ @OldPeter: Feel free to answer it still. :) $\endgroup$ Commented Jan 27, 2017 at 16:34
  • $\begingroup$ @Tito Piezas III I hate to ask, but is there and chance of a typo in your identity? I’ve tried substituting some small solutions, but the RHS of (2) seems vastly too small. Perhaps I’ve misunderstood the notation; I’ve take then RHS of (2) as the product of 20*(x_1^2-y1_^2)(x_1^2-y_2^2)(x_1^2-y3_^2)$(x_2^2-y_12^2)(x_2^2-y_2^2)(x_2^2-y_3^2)(x3_^2-y_1^2)(x_3^2-y_2^2)(x_3^2-y_3^2) $\endgroup$
    – Old Peter
    Commented Jan 30, 2017 at 20:13

2 Answers 2


Define $$R_n=(a_1^n+a_2^n+a_3^n -b_1^n-b_2^n-b_3^n)/n$$ I found and proved that $$R_1S+R_3^3-2R_2R_3R_4+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$ where $S$ is a polynomial containing 9 items.

If $R_1=R_2=0$, then $$R_3^3=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$ Example: $2^n+13^n+21^n=6^n+7^n+23^n$, $(n=1,2)$

If $R_1=R_3=0$, then $$R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$ Example: $2^n+10^n+12^n=3^n+8^n+13^n$, $(n=1,3)$

If $R_1=R_4=0$, then $$R_3^3+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$ Example: $3^n+25^n+38^n=7^n+20^n+39^n$, $(n=1,4)$

If $R_1=R_5=0$, then $$R_3^3-2R_2R_3R_4=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$ Example: $3^n+54^n+62^n=24^n+28^n+67^n$, $(n=1,5)$

Let $a_i=x_i^2,b_i=y_i^2$, $(i=1,2,3)$, the system $(1)$ satisfies $R_1=R_3=0$, then we can prove equation $(2)$.

For more identities of the similar form, please refer to my website on Algebraic Identities

For more examples of numerical solutions, please refer to my website on Equal Sums of Like Powers


You want to prove equation $(2).$ Using equation $(1)$ with $k=2$ solve the quadratic for say $y_3$. Substitute this value of $y_3$ into LHS-RHS of equation $(1)$ with $k=6$ to get a homogeneous $6$th degree polynomial in the $5$ other variables. Substitute the value of $y_3$ into LHS-RHS of equation $(2)$ and verify that the result is a homogeneous $18$th degree polynomial divisible by the homogeneous $6$th degree polynomial.


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